What is the extreme f(x,y)=x^2+2y^2-xy+14y ?

The extreme f(x,y)=x^2+2y^2-xy+14y is Minimum(-2,-4)

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extreme f(x,y)=x^2+2y^2-xy+14y

extreme points

f (x, y) = x^{2} + 2 y^{2} - x y + 14 y

Minimum (- 2, - 4)

Solution steps

Find the critical points:

(- 2, - 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 7

Check the sign of at each critical point

Check critical point

(- 2, - 4) :

Minimum

Minimum (- 2, - 4)

f (x) = - x^{\frac{2}{3}} (x - 2), - 2 \leq x \leq 2

f (x) = - x^{3} + 12 x, (- 6, 5)

f (x, y) = 6 - x^{4} + 2 x^{2} - y^{2}

f (x, y) = 8 y^{2} + x^{2} - x^{2} y

f (x) = - \frac{5}{x - 6}

What is the extreme f(x,y)=x^2+2y^2-xy+14y ?
The extreme f(x,y)=x^2+2y^2-xy+14y is Minimum(-2,-4)